physics hand book

84
Q1.A glass rod rubbed with silk is found to have positive charge of 3.2 x 10 -10 C. a) In this method charge is produced by ........... b) Name the Scientist who establised the types of charges c) Which priciple helps to find the number of electrons transfered. state it. d) Find the number of excess electrons in silk. e) Calculate the mass transfered from glass to silk. Ans: a) Friction b) Benjamin Franklin c) Quantisation of charge. It states that total charge on a body is an integral multiple of fundamental charge, ie. charge of an electron. q = + ne n = 1, 2, 3 ... e = 1.6 x 10 -19 c d) q = 3.2 x 10 -10 c q = ne n = = = 2 x 10 9 electrons e) m e = 9.1 x 10 -31 mass transfered = m e x n = 9.1 x 10 -31 x 2 x 10 9 = 18.2 x 10 -22 kg = 1.82 x 10 -21 kg Q2.Aeroplanes are provided with special rubber tyres that are slightly conducting. why? Ans. By doing so staic charges produced by the friction between the tyre and the road when the aeroplane move through the runway may flow to the earth. Q3.Electostatic experiments are not conducted in humid days. Why? Ans. This is because wet air is slightly conducting and the static charges may get conducted away. 1 & 2. ELECTRIC CHANGES AND FIELDS ELECTROSTATIC POTENTIAL AND CAPACITANCE q e 3.2 x 10 -10 1.6 x 10 -19 Q4.Vehicles carrying inflammable materials usually have metallic chains touching the ground. Why? Ans.Static charges may be produced in the vehicles by friction between the tyre and the road, between the inglammable material and the body of vehicle and also due to air. By doing so the accumulated charges may flow to the earth. Q5.Two charged spheres 3µc and 12µc are seperated by a distance 2 cm apart in air. a) Name the law which governs the force between the charges. State it. b) Find the magnitude force between the charges. c) If the charges are placed in mica (k = 5.6) do the force change. Find the new force. d) Nacl easily converts into ions when put in water. Why? e) What is the dielectric constant of copper? Ans.a) Coulomb’s law (Statement) b) q 1 = 3µc = 3 x 10 -6 c q 2 = 12µc = 12 x 10 -6 c r = 2cm = 2 x 10 -2 m F = = 9 x 10 9 = 9 x 10 9 x = 81 x 10 1 = 810 N c) In a medium the force between the charges decreases by an amount K = ε r F m = = = 14.28 N d) When put in water, the force between Na + and Cl - ions dereases by an amount 81 (ε r of water = 81). So they easily seperates. m 1 4πε 0 q 1 q 2 r 2 q 1 q 2 r 2 3 x 10 -6 x 12 x 10 -6 (2 x 10 -2 ) 2 9 x 10 9 q 1 q 2 k r 2 810 5.6 Aswathy Books 5

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  • Q1.A glass rod rubbed with silk is found tohave positive charge of 3.2 x 10-10C.

    a) In this method charge is produced by...........

    b) Name the Scientist who establised thetypes of charges

    c) Which priciple helps to find the numberof electrons transfered. state it.

    d) Find the number of excess electrons insilk.

    e) Calculate the mass transfered fromglass to silk.

    Ans: a) Frictionb) Benjamin Franklinc) Quantisation of charge. It states that

    total charge on a body is an integralmultiple of fundamental charge, ie.charge of an electron.q = + nen = 1, 2, 3 ... e = 1.6 x 10-19c

    d) q = 3.2 x 10-10cq = ne

    n =

    = = 2 x 109 electrons

    e) me = 9.1 x 10-31

    mass transfered= me x n= 9.1 x 10-31 x 2 x 109= 18.2 x 10-22 kg= 1.82 x 10-21 kg

    Q2.Aeroplanes are provided with specialrubber tyres that are slightly conducting.why?

    Ans. By doing so staic charges producedby the friction between the tyre and theroad when the aeroplane move throughthe runway may flow to the earth.

    Q3.Electostatic experiments are notconducted in humid days. Why?

    Ans. This is because wet air is slightlyconducting and the static charges mayget conducted away.

    1 & 2. ELECTRIC CHANGES AND FIELDS ELECTROSTATICPOTENTIAL AND CAPACITANCE

    qe

    3.2 x 10-101.6 x 10-19

    Q4.Vehicles carrying inflammable materialsusually have metallic chains touchingthe ground. Why?

    Ans.Static charges may be produced in thevehicles by friction between the tyre andthe road, between the inglammablematerial and the body of vehicle and alsodue to air. By doing so the accumulatedcharges may flow to the earth.

    Q5.Two charged spheres 3c and 12c areseperated by a distance 2 cm apart inair.

    a) Name the law which governs the forcebetween the charges. State it.

    b) Find the magnitude force between thecharges.

    c) If the charges are placed in mica (k =5.6) do the force change. Find the newforce.

    d) Nacl easily converts into ions when putin water. Why?

    e) What is the dielectric constant of copper?Ans.a) Coulombs law (Statement)b) q1 = 3c = 3 x 10-6c

    q2 = 12c = 12 x 10-6cr = 2cm = 2 x 10-2m

    F = = 9 x 109

    = 9 x 109 x

    = 81 x 101 = 810 Nc) In a medium the force between the

    charges decreases by an amount K = r

    Fm =

    = = 14.28 N

    d) When put in water, the force betweenNa+ and Cl- ions dereases by an amount81 (r of water = 81). So they easilyseperates. m

    140

    q1q2r2

    q1q2r2

    3 x 10-6 x 12 x 10-6

    (2 x 10-2)2

    9 x 109 q1q2k r2

    8105.6

    Aswathy Books 5

  • e) For metals K = r = (infinity)Q6. Two fixed charges +9e and +4e are

    seperated by a distance d. Whereshould be the third charge placed. Sothat they are in equillibrium.

    Ans. 9e 4 e

    =

    =

    taking squareroot =

    3(x - d) = 2x 3(x-d) = -2x3x - 3d = 2x 3x - 3d = -2x

    x = 3d 5x = 3d

    Which is not possible x = Which is possible

    So the third charge is placed at adistance from the charge 9e.

    Q.7.In HCl molecule H+ & Cl- ions areseparated by a small distance.

    a) Name such an arrangementb) How is it measured. What is its unit?c) What is the net force acting on it when

    placed in uniform Electrifield?d) Derive an expression for torque acting

    on it.e) What happens when it is placed in a

    non uniform electric field?Ans.a) Electric dipole.b) The strength of a dipole is measured

    as dipolemoment. Its unit is Coulombmetre ( c-m)

    c) Zero, net force on eletric dipole is zeroin uniform E.F.

    d) (Derivation of = Ep Sin from text)e) In a non uniform Electric field there is

    an unbalanced External force. So therewill be translatory motion in addition torotation.

    Q.8. Consider a uniform electric field E =3 x 103 i N/C

    q<

    >

    > Pds E

    101

    0

    1010

    140

    q

    r2

    ii) At a point inside the shell the Gaussiandoes not enclose any charge. HenceEF = 0.

    d) Imagine the given square is one of thefaces of a cube of side 10cm.

    = x= x

    = 1.88x105 Nm+2C-1Q.6. Beside its use in storing charge a

    capacitor is a key element of most accircuits.

    a) What are the factors on whichcapacitance of a parallel plate aircapacitor depends?

    b) Define the term dielectric constant of amedium.

    c) Two capacitors C1 and C2 are connectedin series. Derive an expression forcapacitance of the combination.

    d) Find the equivalent capacitance ofcapacitors given in the network.

    (March 2012)

    a) The capacitance of parallel platecapacitor depends on(i) Area of the plates(ii) Seperation of the plates(iii) Dielecric constant of the medium

    b) It is defined as the ratio of capacitanceof a capacitor with medium as dielectricto the capacitance of the samecapacitor with air as dielectric

    K =

    c) Consider two capacitors of capacitanceC1 & C2 connected in series with a

    16

    q0

    16

    10 x10-6

    8.85 x 10-12(No. of facesof a cube = 6)

    65

    F

    12F

    4F 3FA

    D

    B C

    CmCo

    R 121212

    12

    12

    12

    12

    12

    Aswathy Books 13

  • potential V. The charge on eachcapacitor will be same but potential willbe different.

    V1 = and V2 =

    is Cs is the effective capacitance, then

    V =

    But V = V1 + V2

    = +

    ie. = +

    d) Equivalent capacitance betweenterminals A & B = 4 + 12 = 16FEquivalent capacitance betweenB & C = 3 + 6 = 9F Equivalent capacitance of thenetwork is

    = + +

    = + +

    = =

    C = FQ7.How does atmosphere get charged?Ans: By lightning and thunder storms the

    atmosphere get charged.

    Q8.A glass rod held in hand can becharged by rubbing with silk. Whycannot a metal rod be charged like this.

    Ans:Metals and our body are goodconductors. So charges produced byrubbing a metal are transferred to theearth.

    qC1

    qC2

    qCs

    V1 V2

    V

    C1 C2

    qCs

    qC1

    qC2

    1Cs

    1C1

    1C2

    1C

    116

    19

    16/5

    116

    19

    56

    9 + 16 + 5 x 24144

    145144

    144145

    Q9.An electron and a proton is releasedfrom rest in a uniform electric field E.Which will have greater acceleration.

    Ans:Acceleration a =

    Since Me < Mp, Electron has largeracceleration.

    Q10.What is the net force acting on a dipoleplaced in (i) uniform E.F and (ii) Nonuniform Electric field.

    Ans:(1) Net force is zero as the oppositecharges are experiencing a force equalin magnitude but opposite in direction.

    (ii) There is a net force as the force on theopposite charges are not equal andopposite.

    Q11.The electric field is zero at a pointmidway between two equal like charges.What is the potential at that point.

    Ans:The potential is twice the value ofpotential due to a single charge at thatpoint.

    Q12.Can electric potential at any point inspace be zero while intensity of EF atthat point is not zero?

    Ans:At the centre of a dipole there is a fieldbut potential will be zero.

    Q13.The electric field near the surface ofearth is about 100Vm-1. Yet we do notget electric shock. Why?

    Ans:Our body and the earths surface aresituated on an equipotential surface. Sothere is no p.d and hence no shock.

    Q14.What is the work done in rotating adipole once completely.

    Ans:Let be the orientation of a dipole witha uniform E.F. 1 = when the dipole isrotated once completely 2 = + 360.W = PE (Cos1 - Cos2) = PE (Cos - Cos (+ 360)) = PE (Cos - Cos ) = 0

    Q15.A charge is kept at the centre of a circle.Another charge is taken (i) along thecircle (ii) along the ellipse, once

    Eem

    Aswathy Books 14

  • completely. In which case the work doneis more.

    Ans:As the second charge is taken a closedtrip the work done will be zero in boththe cases.

    Q16. If a conducting metal plate ofthickness t is introduced between theplates of a parallel plate capacitor whatchange is produced in the capacitance.

    Ans:The introduction of the conductingmetal plate will decrease the effectivewidth of the dielectric medium to (d - t).Hence the capacitance will increase.

    Q17.Four charges, each equal to q, areplaced at the four vertices of a regularpentagon. The distance of each cornerfrom the centre is a. Find the E.F at thecentre of the pentagon.

    Ans:Imagine a charge q at the corner Ealso. By symmetry, the field at the centreO will be zero. We conclude from herethat the field due to charges at A, B, Cand D is equal and opposite to theelectric field at O due to the charge q.

    So the required EF = along

    OE.

    Q18. A regular hexagon of side 10cm hasa charge 5C at each of its vertices.Calculate the potential at the centre ofthe hexagon.

    Ans.

    140

    qr2

    A B

    E

    D

    CO

    From figure all the six triangles areequilateral OA = OB = OC = OD = OE =OF = r = 10 cm = 0.1 m Potential at thecentre of teh hexagon.

    V = 6 x

    =

    = 2.7 x 106 V

    Q19. Two spheres of copper of same radic,one hollow and the other solid arecharged to the same potential. On whichspere there is more charge?

    Ans. Both would carry equal charge this isbecause the capacity of the spheredepends upon the radius. It does notmatter whether it is hollow or solid.

    Q20. An electric dipole of dipolemoment20x10-6cm is enclosed by a closedsurface. What is the net flux coming outof the surface.

    Ans. Zero. This is because net charge of adipole is zero.

    Q21. How the mass of a body is affected oncharging?

    Ans. When a body is negatively charged itgains mass and when a body is positivelycharged it loses mass.

    140

    qr

    6 x 9 x 109 x 5 x 10-6

    0.1

    )

    ) )600 600600

    F C

    A B

    O

    E D

    Aswathy Books 15

  • 3. CURRENT ELECTRICITY

    Q1. A wire is carrying a currrent. Is itcharged?

    Ans: No. This is because at any time, thenumber of protons in the wire is equalto the number of electrons.

    Q2.Determine the dimensional formula forresistance

    Ans: R = =

    = = ML2T-3A-2

    Q3.Three conductors are connected inparallel. Their conductances are G1, G2& G3. What is their equivalentconductance.

    Ans: = + +

    Equivalent conductance Gp = G1+G2+G3Q4.How does the bend in a wire affects

    the electric resistance.Ans:Since electrons are extremely small in

    size compared to the bends they caneasily change their direction of motion.Thus the resistance of the wire remainunchanged.

    Q5.A 8 resistance wire is bent at themiddle by 1800. Both halves are twistedtogether. What is the new resistance.

    Ans:

    Since length is halved resistance ofeach part is 4.Effective resistance is R =

    R = = 2

    Q6.Give the equivalent of VA-1Ans: Ohm

    Q7.The sequence of colour bands on aresistance is green, yellow, silver. Whatis the resistance?

    Ans:54x10-2. When silver is used third

    VI

    WorkCharge x CurrentML2T-2AT. A

    1Rp

    1R1

    1R3

    1R2

    R1R2R1+R24 x 4

    4+2

    8 4

    4

    colour band, it indicates the multiple of10-2.

    Q8.Resistivity of copper silver andmanganin are 1.7x10-8m, and 44x10-8respectively. Which of these is the bestconductor?

    Ans:Out of these metals, specific resistanceis least for silver. So silver is the bestconductor.

    Q9.Why constantan and manganin used formaking standard resistances?

    Ans:Because of their low temperature co-efficient of resistance and high resistivity.

    Q10.V-I graphs for parallel and seriescombination of two metallic resistors asshown in fig. Which graph representsparallel combination? Justify youranswer.

    Ans:Resistance is less for the combinationA. So A represent the parallelcombination.

    Q11.Two wires of equal lengths, one ofcopper and the other of manganin, havesame resistance. Which wire is thicker?

    Ans:Since two wires are of equal length andequal resistance.R1 = R2

    1 = 2

    =

    Since resistivity 1 of Cu is less thanthat of manganin (2), area of crosssection of Cu (A1) is less than the crosssectional area (A2) for manganin. So

    V

    I

    A

    B

    lA1

    lA2

    A1A2

    12

    Aswathy Books 16

  • manganin wire is thicker than copperwire.

    Q12.Two wires A and B are of the samemetal and of same length. Their areaof cross sections are in the ratio of 2:1.If the same potential difference isapplied across each turn, What will beratio of currents flowing in A & B.

    Ans: I =

    I = =

    In the given problem V, l and areconstantsI A

    = =

    So the ratio of currents are 2:1.

    Q13. What happens to the drift velocity ofelectrons and to the resistance if thelength of the conductor is doubledkeeping potential difference unchanged?

    Ans: vd = =

    Keeping v constant if l is doubled, vd ishalved. Again keeping cross sectionalarea constant, if length is doubled, Rwill be doubled [R l ].

    Q14.Are the paths of electrons straightlines between successive collisions inthe a) Absence of E.F b) Presence ofE.F.

    Ans:a) In the absence of EF paths arestraight lines.

    b) In the presence of EF, the path aregenerally curved due to drif of the E.F.

    Q15.How will you represent a resistanceof 3700 + 10% by colour code?

    Ans: Orange violet red and silver.

    Q16.A wire of resistance 8 is bent in theform of a circle. What is the effectiveresistance between the ends of thediameter of this circle.

    V

    Vl /A

    VAl

    I 1I 2

    A1A2

    21

    evml

    eEm

    Ans: (Same as Question No.5).

    Q17.Two resistances are in the ratio 1:4. Ifthese are connected in parallel, their totalresistance become 20. Calculate thevalue of each resistance.

    Ans: =

    R2 = 4R1

    For parallel combination Rp =

    20 =

    20 =

    R1 =

    = 25R2 = 4R1 = 4 x 25 = 100

    Q18.Why is it not advisable to use Copperwire in a potentiometer?

    Ans:This is because the resistivity of Copperis small and its temperature co-efficientof resistance is large.

    Q19.It is possible to generate 100000 voltpotential difference by rubbing a pocketcomb with wool. Why is it notdangerous?

    Ans:This is because comb (insulator) hashigh resistance. So current is extremelysmall.

    Q20.Would a galvanometer show anycurrent if the galvanometer and cell areinterchanged at the balance point of thebridge?

    Ans:No, The galvanometer will not showany current.

    Q21.Why is it easier to start a car engineon a warm day than on a chilly day?

    Ans:The internal resistance of a car batterydecreases with increase in temperature.

    R1R2

    14

    R1R2R1+R2R1 x 4R1

    R1+4R1

    4R12

    5R120 x 5

    4

    Aswathy Books 17

  • Q22.Why light from a bathroom bulb getdimmer for a moment when the geyseris switched on?

    Ans:The large current drawn from thesupply initially lowers the voltage for amoment until it is stabilised by thetransmission grid.

    Q23.Why is a potentiometer of longer wireconsidered more accurate than apotentiometer of shorter wire?

    Ans:In longer wire the fall of potential perunit length is small. ie Potential gradientis small. Lesser the potential gradient,more accurate is the potentiometer.

    Q24.Can you measure the internalresistance of a car battery with the helpof a wheat stones bridge?

    Ans:The internal resistance of theaccumulator of the car battery is verysmall. Such small resistance cannot bemeasured by a wheatstones bridgesince it become insensitive.

    Q25.Why is voltmeter less accurate inmeasuring emf than a potentiometer?

    Ans:Potentiometer measures potential bynull deflection method. So it will notdraw any current from the source of emfto be measured voltmeter draws somecurrent and consequently measures aslightly lesser value of emf.

    Q26.The circuit in the figure shows use ofa potentiometer to measure the internalresistance of a cell.

    a) If the key is open how does the balancepoint change if the current from thedriver cell decreases.

    b) When the key k is closed, how doesthe balance point change if R is

    increased keeping current from the drivercell constant?

    Ans:(a) Due to decrease in current, thepotential gradient is decreased. Thebalance point would now be obtained ona larger length of the wire.

    (b) If R is increased, the amount of currentdrawn from the cell of emf El willdecrease. The terminal voltage acrossthe cell shall increase. The balance pointwill now be obtained on a larger lengthof the potentiometer wire.

    Q27.What is the magnitude of the resistancein the circuit shown, when no currentflows through the 5 resistance?

    Ans:It is a case of balanced wheat stonesbridge

    =

    X = = 6

    Q28.Calculate the resistance between A andB of the given network.

    Ans: This circuit can be redrawn as

    Here =

    X18

    26

    2 x 186

    K

    E r

    JEl

    G

    R

    vvv

    vvv

    vvvvvv

    X 18

    562

    6V

    vvv

    vvv

    vvv

    vvv

    vvv

    2

    1 10 4 BA

    2

    vvv vvv

    vvv

    vvv

    vvv

    24

    A B10

    1 2

    24

    12

    Aswathy Books 18

  • This represent a balanced wheat stonesbridge. Hence 10 can be neglected.

    Equivalent resistance between A and B is

    R = = 2

    Q29.In the potentiometer circuit shownbelow the balancing point is at X. Statewith reason where the balance point isshifted when

    a) when resistance R is increasing keepingall parameter unchanged.

    b) resistance S is increased keeping Rconstant.

    c) cell P is replaced by another cell whoseemf is lower than that of the cell Q.

    Ans:(a) When R is increased, the currentfrom the driver cell is decreased. Thisdecreases the potential gradient. Sothe balance point will shift to the right.

    b) Increasing resistance s shall have noeffect on balancing point. This isbecause there will be no flow of currentin this branch when potentiometer isbalanced.

    c) The balancing point will not beobtained. For getting balance point onthe wire, the emf of the driver cellshould be greater than the emf of thecell Q.

    Q30.R1, R2 and R3 are different values ofR. A, B and C are the null pointsobtained corresponding to R1, R2 andR3 respectively. For which resistor, thevalue of X will be the most accurate andWhy?

    vvvvvv

    vvv vvv A B

    1

    2 4

    2

    vvvv

    vvvv

    A B

    6

    3

    6 x 36+3

    P R

    JA

    Gvvvv

    S

    B

    Ans:The bridge is most sensitive when thefour resistances of the bridge are ofnearly same magnitude. Hence the nullpoint B will give most accurate value ofX.

    Q31a) Calculate the current through eachcell and 6 resistance.

    b) Find the terminal voltage across E1.

    Ans:

    Applying Kirchoffs second law in mesh 1-3I1 + 4 - 5I1 - 6(I1 + I2) - 4I1 = 0-3I1 - 5I1 - 6I1 - 6I2 - 4I1 = -4

    -18I1 - 6I2 = -49I1 + 3I2 = 2 (1)

    For second mesh-4I2 - I2 + 2 - 5I2 - 6(I1 + I2) = 0

    Rvvvv

    A C

    B

    E1 = 4v

    vvv

    vvv

    vvv

    vvv

    vvv

    5

    5

    4

    4

    r1 = 3

    r2 = 1

    6

    E2 = 2v

    vvv

    vvv

    vvv

    vvv

    vvv

    vvv

    vvv

    2

    1

    I1

    5

    5

    I1 4v I13

    6

    I1 + I2 I1 + I24

    4

    12v

    I2

    Aswathy Books 19

  • -4I2 - I2 + 2 - 5I2 - 6(I1 + 6I2) = -2-16I2 - 6I1 = 2-16I2 +6I1 = 23I1 +8I2 = 1 (2)

    From (1) 9I1 +3I2 = 2From (2) 3I1 +8I2 = 1(2) x (3) 9I1 +24I2 = 3 (3)(1) - (3) -21I2 = -1

    I2 = = 0.0476A

    From (2) 3I1 = 1 - 8I2 3I1 = 1 - 8 x .0476 = 0.6192

    I1 = = 0.206A

    Current through 4v cell = .206ACurrent through 2v cell = 0.0476 ACurrent through 6 resistance = I1 + I2

    = 0.254APotential difference across 6 resistance

    = 6 x 0.254A = 1.52vTerminal potential voltage across

    E1 = IR = E1 - I1r = 4 - 0.206 x 3 = 4 - .62 = 3.38v

    Important Questions

    Q1. Two 120v light bulbs, one of 25w andother of 200w are connected in seriesacross a 240v line. One bulb burnt outalmost instantaneously which one wasburnt and why?

    Ans:As P = , 25w bulb has moreresistance. In series same current glowthrough them. So the 25w bulb willdevelop more heat and burnt outinstanteously.

    Q2.A wire of resistivity is stretched twiceits length. What will be new resistivity.

    Ans:Resistivity will not change as it isindependent of length or area.

    121

    0.61923

    V2R

    Q3.If the length of a wire conductor isdoubled by stretching it keeping thepotential difference across it constant bywhat factor does the drift speed ofelectrons change?

    Ans:Drift speed is related to p.d, v as

    Vd = = ,At constant v, when length is doubled,vd is halved.

    Q4.Why cant we use eight ordinary cellsconnected in series to get a 12v supplyin a car instead battery having 12v.In eight ordinary cells connected inseries the internal resistance will be highand current will be low.

    Q5.Using metre bridge, it is adviced toobtain null point in the middle of the wire.Why?

    Ans: For metre bridge to be more sensitive,its four resistances must be equal. Thisis so when the null point lies near themiddle of the wire.

    Q6.How can we increase the sensitivity ofa potentiometer?

    Ans:By reducing the potental gradient.Potential gradient can be reduced by (i)increasing the length of the wire (ii) byreducing current in the main circuit.

    Q7.What happens if the galvanometer andcell are interchanged at the balancepoint of the bridge?

    Ans:Conditions for balanced bridge remainssame. Galvanometer will show nocurrent.

    Questions from Previous QuestionPapersQ1. A potentiometer wire PQ of length 1m

    is connected to a standard cell of emfE1. Another cell E2 of 1.02 volt isconnected to a parallel combination ofresistance r and a key as shown whenthe switch s is open the null point isobtained at a distance of 51cm from P.

    a) What do you mean by potential gradient.b) Determine the potential gradient across

    evml

    eEm

    Aswathy Books 20

  • PQ.c) When the switch s is closed, will the

    null point move towards P or Q? Justifyyour answer? (March 2010)

    Ans:a) Potential gradient is defined as thepotential per unit length.

    b) Potential gradient across PQ == E1 v/m

    c) The null point does not change. Theresistance connected in series togalvanometer does not change the nullpoint.

    Q2.

    a) Find the equivalent resistance of abovecircuit.

    b) Determine the current through the 75resistance. (March 2011)

    Ans:a) Equivalent resistance of parallelcombination is

    = + + = +

    =

    =

    R = = 18.75

    Effective resistance of the combinationis = 100 + 18.75 = 118.75

    b)

    E11

    vvv

    1.02v

    E1

    P rJ

    G

    s

    vvv

    vvv

    vvvvvv

    100

    6v50

    7550

    1R

    175

    150

    150

    250

    175

    3x2 + 1x2150

    1R

    8150

    1508

    vvv

    vvvvvv

    vvv

    .05A106v

    I1

    50 75

    (.05-71)

    50

    I1

    Effective current = = .05A

    Voltage across 75 = voltage across 5075 r (.05 - 2I1) = 50I13.75 - 150I1 = 50I13.75 = 200I1

    I1 = = .01875

    Current through 75 resistance= .05 - 2 x .01875= .05 - .0375= .0125A

    Q3.Remya makes the following circuit tomeasure the emf of a cell.

    She says that the voltmeter reading willgive the emf of the cell.

    a) The physics teacher says that it is notpossible to measure emf of the cell inthis way. Justify this statement.

    b) Explain with a circuit diagram the methodto measure the emf of a cell with apotentiometer if you are given anotherstandard cell (whose emf is known)

    (March 2011)Ans:a) Emf cannot be measure in this way

    since the cell is connected to aresistance. Emf is the potential differenceacross the terminals of a cell in a closedcircuit.

    b)

    With the given cell let l1 be the balancinglength, then E l1With the standard cell E1 let l2 be thebalancing length,

    6118.75

    3.75200

    vvv

    E

    R

    v

    vvv

    ( )

    GE

    B

    A+

    -

    + -

    J

    Aswathy Books 21

  • then E1 l2

    =

    E = x E1

    Since E1 is known, E can be measured.

    Q4.Thermal velocity of electron is about106m/s and drift velocity is about10-3m/s.

    a) What is mean by drift velocity.b) Obtain the relation between drift velocity

    and electric current (March 2011)Ans:a) The velocity with which the

    electrones are drifted under theinfluence of an electric field is called driftvelocity.

    b)

    Consider a conductor of length l, crosssectional area a carrying n electronsper unit volume. When an electric fieldE is applied let v be the drift velocity.Volume of the conductor = alNumber of electrons in the conductor =nalTotal charge in the conductor = naleTime taken by the electrons to flow fromB to A, t = l/vTherefore current i = q/t

    =

    i = nave

    Q5. n equal resistances, each of valueR are combined in different ways.

    a) What are possible values of maximumand minimum equivalent resistances?

    b) If the maximum and minimum valuesare 40 and 2.5 respectively. Find nand R. (March 2008)

    Ans:a) Maximum resistance is obtainedwhen connected in series

    EE1

    l1l2l1l2

    EV

    A Bl

    nalel/v

    Rs = R + R + .... n timesRs = nRMinimum resistance is obtained whenconnected in parallel.

    = + ... n times

    =

    Rp =

    b) nR = 40 (1)

    = 2.5 (2)

    (1)/(2)

    n2 = 16 ie n= 4

    R = = 10Q6.An electric kettle has two heating coils.

    When one of the coils is switched on thekettle begins to boil in 6 minutes andwhen the other is switched on the boilingbegins in 8 minutes.

    a)The working of an electric kettle is basedon which principle? State the principle.

    b) Find the ratio of the resistances of thetwo coils in the above problem.

    c) If the two coils are connected in series,find in what time will the boiling begin forthe same quantity of water.

    d) Suggest a method to reduce the boilingtime. Find the boiling time then.

    (March 2009)Ans:a) Joules law. The amount of heat

    developed in a conductor by passage ofa steady current through it is directlyproportional to the square of the current,resistance of the conductor and time offlow.

    b) Let Q be the heat required to boil thekettle and v be the supply voltage.

    Q = r t1 = r t2

    = = =

    c) When connected in series let t be thetime taken

    1Rp

    1R

    1Rp

    1R

    nR

    Rn

    Rn

    402.5

    404

    v2R1

    v2R2

    R1R2

    t1t2

    68

    34

    v

    Aswathy Books 22

  • Q = x t

    R1 + R2 = R2 + R2 = R2

    Q = (1)

    d) Time t = 6 + 8 = 14 minutesWhen the coils are connected in series,the equivalent resistance increases andhence more time is required. Whenconnected in parallel, equivalentresistance becomes.

    = + = +

    R = =

    = R2

    Let t be the time taken in this case

    Then Q = x T (2)

    From (1) & (2) x t = x T

    =

    T = = 5minutes

    Q7a) The resistance R of a conductordepends on its length l area of crosssection A and resistivity of the material. The correct expression connectingR, l, A and is

    (i) R = (ii) R= (iii) R =

    (iv) R = Alb) The voltage-current graphs for two

    resistors of the same material andsame radii with length L1 and L2 areshown in the figure. If L1 > L2 state thereason, which of these graphs

    v2(R1 + R2)

    34

    74

    v2t7/4 R2

    1R

    1R1

    1R2

    17/4R2

    1R2

    7/4R2 x R27/4R2 + R2

    7R2

    4 + 114711

    V2

    711

    R2

    V2

    74

    R2

    V2

    711

    R2

    t1.75

    T0.63

    14 x .631.75

    Al

    Al

    lA

    represent voltage change for L1?(March 2012)

    Ans:a) (iii) R =

    b) B represents voltage change for L1.Resistance of B is greater than that forA, since resistance increases with length.

    Q8.To determine the resistance of a wire astudent uses a metre bridge.

    a) Draw the circuit diagram and explain theworking of a metre bridge to determinethe resistance of a wire.

    b) How the resistance of a wire changes ifits (i) temperature (ii) length is increased.Explain. (March 2008)

    Ans:

    Principle: The Wheatstones bridgesbalancing condition.Uses: Metre bridge is used to find theunknown resistance, resistivity etc.Experiment: Connections are made asshown in the figure. Circuit is closed andsuitable resistance R is introduced in theresistance box. The jockey is moved alongthe wire AB and balancing length ismeasured at the null point, AJ = l cm, BJ =(100-l )cm. Let r be the resistance per unitlength of the wire.According to Wheatstones bridge principle

    =

    X =

    Knowing R & l, unknown resistance X canbe determined.

    V

    I

    A

    B

    lA

    vvv GR

    BA J

    X

    l r(100 - l )r

    Rl(100 - l )

    XR

    Aswathy Books 23

  • Q8.A wire is drawn to double its originallength. What will be increase in its (i)resistance (ii) resistivity.

    Ans: R =

    When length is doubled, area will behalved

    Rl = = = 4 = 4R

    ie the resistance increases.Increase in resistance = 4R - R = 3R.

    (ii) Resistivity will not change as it isindependent of the dimensions of thewire.

    Q9.Explain why Aluminium wires arepreferred for overhead power cables?

    Ans:Aluminium wire is preferred becausei) It has low density (lighter) (ii) low

    resistance(iii) It is cheaper compared to copper.

    Q10.n equal resistors are connected inseries, then in parallel. Find the ratio ofseries to parallel combination.

    Ans:In series Rs = R + R + ... n times Rs = nR

    In parallel, = + + .. n times

    =

    Rp =

    Rs: Rp = = = n2:1

    Q11.Explain why Copper is not used inpotentiometer

    Ans: Resistance of Copper wire is verysmall. Hence when Copper is used forthe potentiometer wire, it is as if shortcircuited. No p.d will be available on thewire.

    Q12. Is the dimensional formula ofelectromotive force same as that offorce?

    Ans. No, the dimension formula of emf issame as that of potential. (work/charge)

    lA

    l lAl

    (2l )A/2

    lA

    1Rp

    1Rp

    nR

    1R

    1R

    Rn

    nRR/n

    n2

    1

    Q13.Why the jockey should not be rubbedagainst the potentiometer wire?

    Ans. Because rubbing can affect theuniformity of the cross-sectional area ofthe wire.

    Q14. A wire of resistivity is stretched tothere times its length. What will be thenew resistivity?

    Ans. Resistivity remains constant since it isindependent of the dimension of thematerial.

    Q15.Electric current is the rate of flow ofcharges through any cross section of aconductor.

    a) What do you meant by current density.What is its unit?

    b) Obtain the relation between currentdensity and conductivity?

    Ans:a) The electric charge flowing normallyper unit area per second through aconductor is called current density. Itsunit is Am-2.

    b) Consider a conductor of lengthl, area ofcross section A, given a potentialdiffernce V. Let E be the electric field andI be the current produced.By Ohms law V I

    V = RI

    V = x I

    But = j, current density

    V = l jBut potential V = E x lEl = l jE = jj =

    But = , conductivity

    j = E

    This is the relation between current,density and conductivity.

    Aswathy Books 24

    lA

    IA

    1

    E

  • 4. MAGNETIC EFFECTS OF CURRENT AND MAGNETISM

    Q1.Which of the three has lowestresistance - Ammeter, voltmeter,galvanometer?

    Ans:Ammeter.(An ideal ammeter has zero resistanceand an ideal voltmeter has infiniteresistance).

    Q2.A negative charge is moving verticallydownwards when it enters a magneticfield directed to the south, what is thedirection of force on the charge.

    Ans:East direction (Applying Flemings lefthand rule).

    Q3.How can we increase the range of (i)an ammeter (ii) a voltmeter.

    Ans:(i)The range of an ammeter can beincreased by connecting a lowresistance parallel to it.

    (ii) Range of voltmeter can be increasedby connecting a high resistance inseries with it.

    Q4.Why do we prefer phosphor-bronzealloy for the suspension wire of amoving coil galvanometer.

    Ans:Phosphor bronze fibre has smallrestoring torque per unit twist and ahigh tensile strength.

    Q5.A charged particle entering at an angle850 with the magnetic field. Its path willbe ..........(Helical)

    Q6.Consider the circuit shown in the figurewhere APB and AQB are semicircles.What is the magnetic field at the centreC of the loop.

    Ans: Magnetic fields dueto the two semicircles areequal in magnitudebut opposite in direction.Moreover the M.F at Cdue to the striaght parts is zero. So thenet M.F at C is zero.

    P

    A B

    Q

    Q7. Why an ammeter is always connectedin series?

    Ans: Ammeter is a low resistanceinstrument. When it is connected inseries in a circuit it does not appreciablychange the resistance of the circuit.Consquently the current to be measuredis not appreciably changed.

    Q8.A straight section AB of a circuit liesalong the X axis from x = -a/2 to x = +

    a/2and carries a steady current I. What isthe M.F due to the section AB, at a pointx = +a.

    Ans:Zero. This is because observation pointis on the axis of the straight section (M.Fwill be perpendicular to this line).

    Q9.An particle and an electron areprojected into a uniform M.F, B with thesame velocity v such that v isprependicular to B. Compare the periodof revolution of the particle with that ofproton.

    Ans:If m and q are the mass and charge of

    a proton, then period t =

    For an particle mass = 4m and charge= 2q

    t = =

    = 2 x time period of a proton.

    Q10.Establish the dimensional formula forB.

    Ans: F = Bqvsin

    B = =

    ( has no dimension)= MT-2A-1

    Q11.What is the MF a the centre of a currentcarrying cube made of twelve wires?

    2mBq

    2(4m)B(2q)

    2 x2mBq

    [F][q] [v]

    MLT-2

    ATLT-1

    Aswathy Books 25

  • Ans:

    The given system may be regarded asa set of six current carrying pairs. Thecontribution of each pair is zero. So thenet MF induction at the centre is zero.

    Q12.Apply Amperes law qualitatively to thethree parts as shown below.

    Ans:For path I and III, B. dl = 0iFor path II B.dl = 0 since the net currentis zero.

    Q13.What do you mean by the term currentelement. What is its SI unit. What is itssignificance?

    Ans:Current element denoted by idl is theproduct of current and geometricallength of an element underconsideration. Its SI unit is Amperemetre (Am). Just electric charge isconsidered as a source of E.F currentelement is considered as a source ofM.F.

    Q14.If a student by mistake connect avoltmeter in series and ammeter inparallel to a circuit, what will happen?

    Ans:When voltmeter is connected in series,the resistance of the circuit becomeshigh. So the current decreases and thevoltmeter will not read the requiredpotential difference.When the ammeter is connected inparallel in a circuit, the resistance isconsiderably reduced so a large currentwould flow which will damage theammeter.

    0 0

    >>

    > >

    > >>

    >

    > >

    > >

    >

    >

    Q15.An electron and proton moving in samedirection with same kinetic energy entersa perpendicular magnetic field. Will theydescribe paths of same radius.Why?

    Ans:The radius of electron will be smaller.

    Bqv =

    Bq = = =

    r =

    For the given problem r m. Electronbeing lighter shall describe path ofsmaller radius.

    Q16.A rectangular loop of area A having Nturns and carrying a current of I ampereis held uniform in a MF, B.

    a) What is the maximum torqueexperienced by the loop?

    b) In which orientation will the loop be instable equilibrium.

    Ans:a) = NBIA.b) when the plane of the coil is prependicular

    to B it experience no torque and istherefore in stable equilibrium.

    Q17.How will the magnetic field intensity atthe centre of a circular coil carryingcurrent change if the current through thecoil is doubled and the radius of the coilis halved.

    Ans: B =

    Bl = = 4 = 4B

    MF is increased by a factor 4.

    Q18.The energy of a charged particlemoving in a uniform M.F does notchange.Why?

    Ans:As M.F is a conservative field it doesno work in a charged particle. So theenergy of the particle does not change.

    Q19.Two wires of equal lengths are bent inthe form of two loops. One is squareshaped while the other is circular. These

    mv2r

    mvr

    Pr

    2mEkr

    2mEkBq

    0I2r0(2I)2(r/2)

    0I2r

    >

    >

    i

    i

    I

    IIIII

    Aswathy Books 26

  • are placed in a uniform M.F & samecurrent is passed through them. Whichloop will experience greater torque.Give reason.

    Ans:For a given wire circular loop hasgreater area than the square loop.Since the torque experienced by theloop is proportional to the area of theloop, the circular loop experiencegreater torque than square loop.

    Q20.A charge 8c is moving with a velocity

    v = (7j + 4k) in a magnetic field

    B = (7j + 4k) Wbm-2. Find the forceacting on the charge.

    Ans:v and B are parallel vectors.

    v x B = 0

    F = q (v x B) = 0

    Q21.Which one of the two, an ammeter ora milli ammeter, has a higher resistanceand why?

    Ans:Shunt resistance s =

    So the shunt needed to convert agalvanometer into a milliammeter hasa larger value than that required toconvert into ammeter. As the shunt isconnected in parallel with thegalvanometer, the milliammeter willhave higher resistance than theammeter.

    Q22.An electron beam is moving verticallydownwards if it passes through amagnetic field which is directed fromsouth to north in a horizontal plane, thenin which direction the beam would bedeflected?

    Ans:

    The encircled cross represent thedirection of v of electron. Applying righthand rule for the cross product of

    ^ ^^ ^

    Ig x GI - Ig

    X

    S

    W E

    NB

    vectors we find the direction of v x Btowards east. Since Fm = -e (v x B).Therefore the force is directed west.

    Q23.Why a cyclotron is not suitable foraccelerating electrons?

    Ans:When an electron is accelerated in acyclotron, it very soon acquires a veryhigh velocity. Its mass begins to vary as

    m = . The synchronisation

    between the applied frequency andfrequency of revolution of the charge isdisturbed. Thus the cyclotron is notsuitable for accelerating electrons.

    Q24.What are the advantages of soft ironcore used in moving coil galvanometer?

    Ans:a) The magnetic lines of forces crowdthrough the soft iron core. This increasesthe MF and hence the sensitivity of thegalvanometer.

    b) The soft irron core helps to make the MFradial.

    Q25.The wire shown in the figure belowcarries a current of 60A. Determine themagnitude of the magnetic field at C.Given radius = .02m.

    Ans: B = = dl Sin 90

    = dl = x= =

    = 4500 x 10-7T= 4.5 x 10-4T

    Q26.An electron in an atom revolves aroundthe nucleus in an orbit of radius 0.5A0.Calculate the equivalent magneticmoment if the frequency of revolution ofthe electron is 1010MHz.

    Ans:M = iA r = .5A0 = .5 x 10-10m

    m0 1 - v2/c2

    04

    idl Sinr2

    0i4r2

    0i4r2

    0i4r2 x 2r

    34

    30i8r

    3 x 4 x 10-7 x 60 8x .02

    >

    >

    >>

    ) 900

    C

    Aswathy Books 27

  • >>

    >

    >

    R1R2

    i

    i

    = e x r2 = 1010MHz= (1.6 x 10-19) x 1010 x 3.14 x (0.5 x 10-10)2= 1.26 x 10-23 Am-2

    Q27.Determine the M.F at the point O inthe system shown below.

    Ans:MF at the centre of a current carrying

    circle is and so that of a semicircle

    is

    B = B1 + B2

    =

    More Application Questions

    Q1.Write two properties of a material usedas a suspension wire in a moving coilgalvanometer.

    Ans: i) Low value of torsional constant (k)ii) High conductivity

    Q2.An electron and a proton movingparallel to each other in the samedirection with equal momenta enter intoa uniform magnetic field which is at rightangles to their velocities. Trace theirtragectories in the magnetic field.

    Ans: They follow circular tracks of equalradii, electrons revolving in clockwiseand protons in anticlockwise inwardnormal to the plane of the paper.

    Q3.Two wires of equal lengths are bent inthe form of two loops. One of the loopsis square shaped whereas the otherloop is circular. These are suspendedin a uniform magnetic field and samecurrent is passed through them.Whichloop experience greater torque? Givereasons?

    Ans: Torque = i (A x B)As for same perimeter area of the

    0i2R0i

    4R

    1R1

    1R2

    +0i4

    square is more it will experience moretorque.

    Q4.Write one condition when an electriccharge experience no force in amagnetic field.

    Ans:When the charge is moving parallel tothe M.F.

    Q5.Which one of the following will describethe smallest circle when projected withsame velocity v perpendicular to themagnetic field B.

    (i) particle (ii) particleAns: qvB =

    r =

    Since value of q is large for particle, itdescribes smaller circle.

    Q6.An ammeter and a milliammeter areconverted from the same galvanometer.Out of the two which current measuringinstrument has higher resistance.

    Ans:Higher is the range lower will be thevalue of shunt. So milliammeter will behaving higher resistance.

    Q7.What is advantage of using radialmagnetic field in a moving coilgalvanometre.

    Ans: (i) Maximum torque is experienced.(ii) Torque is uniform for all positions ofthe coil.(iii) Plane of the coil is parallel to thedirection of magnetic field.

    Q8.Equal currents are flowing through twoinfinitely long parallel wires. The M.F ata point midway when the currents areflowing in the same direction is

    Ans: (zero)

    Q9.A positive charge moving verticallyupwards enters a magnetic field directedtowards north. In which direction will theforce act on the charge.

    Ans:Force will act towards west.

    Q10.An electric current is flowing due southalong a power line. What is the direction

    mv2r

    mvBq

    O

    Aswathy Books 28

  • of the magnetic field (a) above it (b)below it.

    Ans: a) Towards west above the line b) Towards east below the line

    Q11.A current carrying loop, free to turn isplaced in a uniform magnetic field B.What will be its orientation, relative toB in the equilibrium state?

    Ans:In equilibrium state, the plane of thecurrent carrying loop will be normal toB.

    Previous Questions

    Q1. Oersted found that moving charges orcurrents produce a magnetic field in thesurrounding space.

    a) An electric current is flowing due southalong a power line. What is the directionof MF (i) above it (ii) below it.

    b) Draw a neat and labelled diagram of acyclotron. State the underlying principleof its working.

    c) A cyclotrons oscillator frequency is10MHz. What should be the operatingMF for accelerating protons e = 1.6 x10-19C and Mp = 1.67 x 10-27kg.

    (March 2012)Ans:a) (i) Towards west above the line.(ii) Towards East below the line.b) Working of cyclotron in NCERT text).c) F = 10 x 106Hz

    F =

    B =

    =

    = 0.655T

    Q2a) State Flemmings left hand rule.b) Figure shows the section of a straight

    conducting rod placed in an externalM.F, B. The number density of chargecarrier is n. Deduce the forceexperienced by charge carriers. Whatis the name given to this force? If the

    Bq2m2mf

    q

    2 x 3.14 x 1.67 x 10-27 x 10 x 106

    1.6 x 10-19

    conductor is of an arbitrary shapesuggest a method to find this force.

    (March 2008)

    Ans:a) If the middle finger, forefinger andthumb of the left hand are stretched outmutually perpendicular to each othersuch that the middle finger is in thedirection of the current the forefinger inthe direction of MF then the thumb givesthe direction of the force on theconductor.

    b) Force on the current carrying conductorisF = i ( l x B)But i = nAVdeF = nAVde ( l x B)Al is the volume and neAl is the totalcharge q.F = nAle (Vd x B)

    F = q (Vd x B)This the Lorentz force.For equilibrium, the charges accumulatein the perpendicular direction to the fieldconstituting a Hall voltage. By measuringthe voltage we can find the force.

    Q3.A galvanometer is used to detect currentin a circuit.

    a) State the working principle of agalvanometer.

    b) How will you convert it into (i) anammeter and (ii) a voltmeter.

    c) A galvanometer coil has a resistance of12ohm. It shows a full scale deflectionfor a current of 3mA. How will youconvert this into a voltmeter of range0-18v. (March 2010)

    Ans:a) A current carrying loop when placedin a magnetic field experience a torque.At equilibrium this torque is balanced bythe torsional couple of the suspensionfibre.

    nBiA = c

    l

    vdq A

    > > > > > > >

    B

    Aswathy Books 29

  • b) A galvanometer can be converted to anammeter by connecting a lowresistance called shunt parallel to it.

    Shunt S =

    A galvanometer can be converted to avoltmeter by a connecting a highresistance in series with it

    High resistance R = - G

    c) ig = 3 x 10-3A, G = 12 , v = 18vThe resistance required to convertgalvanometer in the voltmeter.

    R = - G

    = = 5988

    By connecting a resistance R = 5988in series with a galvanometer, we canconvert it into ammeter.

    Q4.Amperes circuital theorem is generallyused to determine the magnetic fieldproduced by a current carrying element.

    a) State Amperes circuital theorem.b) Obtain an expression for infinitely long

    straight conductor using Amperescircuital theorem.

    c) A long straight conductor carries 35A.Find the MF produced due to thisconductor at a point 20cm away fromthe centre of the wire? (March 2010)

    a) According to Amperes circuitaltheorem, line integral to magnetic fieldaround any closed path is 0 times totalcurrent enclosed by the path.

    B. dl = 0 . ib) Consider a solenoid of length l having

    n no. of turns per unit length.

    igGi - ig

    Vig

    Vig

    183 x 10-3

    12

    0

    P

    A B>

    B = 0S R

    O Q

    To find the M.F at the point O image arectangular path PQRS containing thepoint O.

    By the theorem B. dl = 0 x i

    B. dl + B. dl + B. dl + B. dl

    = 0 nliBl + 0 + 0 + 0 = 0nli

    B = 0nic) i = 35A

    d = 20cm = .2m

    B = x

    = 10-7

    = 350 x 10-7 = 3.5 x 10-5 T

    Q5.A student claims that if lightening strikesa metal flag pole, the force exerted bythe earths M.F on the current carryingpole can be large enough to bend it.(Lightning currents are very large of theorder of 104 to 105A).

    a) What is your opinion about the studentsabove argument. Justify the abovestatement.

    b) Which law helps you to find the directionof force on the pole. State the law.

    c) Obtain an expression for force on thepole by taking l as length, B as theearths horizontal intensity and i as thecurrent due to lightning.

    Ans:a) F = Bil Sin = Bil= 0.38 x 10-4 x 105 x 10= 38 N B = 0.38 x 104

    i = 105Al = 10m

    This force is so feeble that the pole willnot bend.

    b) Flemings left hand rule:-According to this rule, stretch the thumb,forefinger and middle finger to representmutually perpendicular directions. If the

    X Y

    0 QP RQ SR PS

    04

    2id

    2 x 35.2

    Aswathy Books 30

  • forefinger points in the direction of thefield and the middle finger points in thedirection of motion of a positivelycharged particle, then the thumb pointsto the direction of force.

    c) Let v be the drift velocity of e be thecharge of one electron.Magnetic Lorentz force f = e (v x B)If n is the number density of freeelectrons, total no. of free electrons inthe conductor N = nAlwhere A area of cross section and l1length of the pole.Total force on the conductor F = Nf

    F = nAl e (v x B)

    = nAl e vBSinBut current i = nAve

    F = Bi l SinF = i (l x B)

    Q6.Two conducting wires AB and CD areconnected to the battery.

    (a) (b)

    a) When they are connected as shown infig (a), what is the force betweenthem.Why?

    b) When connected as in fig (b) what isthe force. Why?

    c) What is the rule governing the abovephenomenon?

    Ans:a) Repulsion. Since the conductors ABand CD carry currents in oppositedirection.

    b) Attraction, since they carry current insame direction.

    c) Flemmings left hand rule.

    Q7.What happens to the field of a toroidwhen the radius is doubled keeping the

    number of turns per unit length same?Ans:When the radius is doubled the field

    remains the same since B = 0niindependent of the radius of the toroid.

    Q7. A Voltmeter is always connected inparallel to a circuit. Why?

    Ans. Resistance of a Voltmeter is very high.So it will draw only a small current whenconnected in parallel.

    Q8. An particle moving in a straight lineenters a uniform magnetic field parallelto the field direction. How will its pathand velocity change.

    Ans. We know that F = Bqu SinHere = 0 F = 0Since force acting on the -particle iszero, the velocity and path will notchange.

    Q9. An electron is moving along a straightpath in a certain region in space. Canwe conclude there is no magnetic fieldin this region?

    Ans. If the electron is moving, parallel tothe direction of the magnetic field, thenit will not experience any force. So inthe given problem we cannot concludethat there is no force in the region.A

    B C

    DA

    B C

    D

    Aswathy Books 31

  • 5. MAGNETISM & MATTER

    Q1.Give three reasons of makingpermanent magnet with alnico.

    Ans.(i) Large area of the hysteris is loop(ii) High coercivity(iii) High retentivity.

    Q2. Horizontal and vertical componentsofearths M.F at a place are 0.22T and0.38T respectively. Find the resultantintensity of earths M.F.

    Ans. B = Bh2 + Br2 = 0.222 + 0.382 = 0.44 T

    Q3. The angle of dip two places on thesurface of the earth are respectively 00and 900. Where are these placeslocated.

    Ans. At the magnetic equator angle of dipis 00. At the magnetic pole, angle ofdip is 900.

    Q4. The horizontal component of MF at aplace 3 times the vertical component.What is the value of angle of dip at theplace.

    tan = = =

    = 300Q5.What is S1 unit of magnetic

    susceptibility?Ans. Tm A-1

    Q6. Why do magnetic lines of force plyerto pass through iron than air?

    Ans. Permeability of soft iron is greaterthan that of air.

    Q7. Does an iron bar magnet retain itsmagnetism when melted? Give reasonfor your answer?

    Ans. Iron melts at a temperature which ishigher than the curie temperature ofiron. So the iron bar magnet cannotretain its magnetism when melted.

    BvBn

    Bv3 Bv

    13

    Q8. Find out the value of Bohr Magneton.Givenh = 6.63 x 10-34 Js & e = 1.6 x 10-19 c

    Ans. 1 Bohr Magneton =

    =

    = 9.27 x 10-24 Am2

    Q9.What happens to a bar magnet if it iscut into two pieces (i) transverse to itslength (ii) along its length.

    Ans. When cut transversely, thepolestrength of each pole is the sameas that of the bar magnet but the lengthis halved. So magnetic moment ishalved.

    When the bar magnet is cut along itslength, the pole strength of each pieceis halved but the length is unchanged.So magnetic moment is halved.

    Q10. Two identical thin bar magnets eachof length L and pole strength m areplaced at right angles to each other withthe N pole of one touching the S- poleof the other. Find the magnetic momentof the system.

    eh4 me

    1.6 x 10-19 x 6.63 x 10-34

    4 x 3.14 x 1.67 x 10-27

    N S

    N S

    L

    N2

    S2

    N1 S1

    ^

    ^

    < >

    Aswathy Books 32

  • Ans. The system bahaves as a magnetwhose pole strength is m and lengthN2S1N2S1 = N1S12 + N2S22 =

    = L2 + L2 = 2 LMagnetic moment = polestrength xlength

    = m 2 L = 2mLQ11. a) Magnetic lines of force represent

    lines of force of a moving chargedparticls at every point.

    b) Magnetic field lines can be entirelyconfined within the core of toroid, butnot within a straight solenoid. Why?

    c) If magnetic monopoles existed, howwould Gausss law of magnetismmodified?

    d) Does a bar magnet exert a torque onitself due to its own field? Does oneelement of a current carrying wire exertforce on another element of the samewire?

    e) Can a system have magnetic momenteven though its net charge is zero?

    Ans. a) Magnetic Lorent =

    foce is F = q ( x B )

    Magnetic force is always normal to

    B .magnetic lines of B cannot representthe lines of force of a moving particle.

    b) Magnetic field lines can be entirelyconfined to the core of a toroid becausea toroid has no ends. It can confinethe field within its core.A straight solenoid has two ends. Ifthe entire magnetic flux were fonfinedbetween these ends, the M.F lines willno longer be continuous.

    c) According to Gausss theorem inmagnetism, magnetic flux over anysurface is always zero ie B . ds = 0If monopoles existed, the magnetic fluxwould no longer be zero, but equal to

    0

    0 times the polestrength enclossed bythe surface.ie B . ds = 0 m

    d) No three is no force or torque on anelement due to the field produced bythat element itsely. But their is a forceon an element of the same wire. Forspecial case of straight wire this forceis zero.

    e) Yes, a system can have magneticmoment even if its net charge is zero.For eg. every atom of para andferromagnetic material has a magneticmoment though atom is electricallynuetral

    Q12. What is the approximate distanceupto which earths MF extends.

    Ans. 3.2 x 104 km

    Q13. What is the basic difference betweenmagnetic and electric lines of force?

    Ans. Magnetic lines of force are closed,continuous but electric lines of force arediscontinuous.

    Q14. If a compass box and dip circle wereto be taken to the magnetic north poleof the earth, what would one observewith regard to directions of theirrespective needles & why?

    Ans. The needle of compass box shall notnecessarily stand along north southdirection. If may point along anyarbitrary direction.

    The needle of the dip circle shallstand vertical with south pole pointingupwards.

    Q15. The susceptibility of magnesium at300k is 1.2 x 10-5. At what temperaturewill be susceptibility be equal to 1.44 x10-5?

    Ans. Xm =

    =

    T| = = = 250K

    0

    CT

    XmX|m

    T|

    TXmX|m

    T 1.2 x 10-5 x 3001.44 x 10-5

    Aswathy Books 33

  • Q16. The angle of dip at a location in southIndia is about 180, would you expect agreater or lower dip angle in Britain?

    Ans. Yes we expect greater dip angle inBritain because it is located close to thenorth pole.

    Q17. In which position, the potential energyof a magnet in a uniform M.F is a) zeroand (ii) Maximum

    Ans. (i) Potential energy of a magnet is zerowhen it is held perpendicular to the M.F.

    (ii) It is maximum when held antiparallel tothe M.F.

    Q18. Why does a paramagnetic substancedisplay greater magnetisation whencooled? How does a diamagneticsubstance respond to similartemperature changes?

    Ans. As temperature decreases the kineticenergy of the dipoles decreases & theybecome more ordered. Soparamagnetism increases.

    Magnetism of a diamagneticsubstance is independent oftemperature changes.

    Q19. You are given two identical barmagnets having differentdipolemoment. How will you select theone having larger dipole moment.

    Ans. The two bar magnets are suspendedin two fibres and their period ofoscillation is calculated. The one havingless period has more magneticmoment.

    Q20. What type of Feromagnetic materialis used for coating magnetic tapes incassette player or for building memoryof a modern computer.

    Ans.Ceramics or ferrites.

    Q21. Three identical specimens ofmagnetic materials Nickel, Antimony,Aluminium are kept in a non-uniformM.F. Draw the modification in the fieldlines in each case. Justify your answer.

    Ans. Antimony is diamagnetic material,permeability is less than unity. So itrepels magnetic lines of force.

    Aluminium is paramagnetic material,permeability is greater than unity. So itattracts magnetic lines of force.

    Nickel is ferromagnetic materialpermeabity is very large in comparisonto the paramagnetic material. Thus ithas greater tendency to attractmagnetic lines of force.

    Q22. The magnetic field lines of the EarthsM.F is from south to north. By theconvention of a magnet, the Earthsmagnetic north is at the geographicsouth. The M.F lines are also slightlyinclined from the north-south direction.

    a) Name the quantities that conventionallyused to specify the earths M.F.

    b) If you made M.F map of the earths fieldat Melborn in Australia the lines seen tocome out of the ground. In whichdirection is a compass needle free tomove?

    c) The earths field roughly estimates thefield due to a dipole of moment 8 x 1022JT-1, check the order of magnitude ofthis field.

    Aswathy Books 34

  • Ans.a) Declination, dip and horizontalintensily are the independent quantitiesused to specify the earths M.F.

    b) At the geometical north or south polethe earths M.F is exactly vertical. So isat Melborn in Australia. A compassneedle free to rotate in a horizontalplane cannot give a proper direction, iecompass needle can move freely.

    c) B = =

    = 0.3 x 10-4 Tr = radius of earth.

    Q23. The hysterisis loop of a soft iron piecehas a much smaller area than of acarbon steel piece. These materials areto go through repeated cycles ofmagnetization.

    a) Which piece will dissipate greater heatenergy.

    b) A system displaying hystensis loop is adevice for storing memory - Explain.

    c) What are the difference betweenpermanent magnets and electromagnets.

    Ans. a) The area of the hysterisis loop isproportional to the heat energy lost percycle of magnetization. Carbon steelpiece will dissipate more heat than softiron peice. Steel has greater area ofhysterisis loop than soft iron.

    b) Magnetization of a ferromagneticdepends on the field and number ofcycles of magnetization it has gonethrough. The valve of magnetization isa record (memory) of its cycles ofmagnetization. If information bils canbe made to corresponding cycles. thesystem displaying such a hysterisis loopcan act as a device for storinginformation.

    c) Permanent magnets have highretentivity and high coercivity. Theyhave low loss of magnetisation. eg.steel.

    04

    mr3

    8 x 1022 x 10-7

    (6.4 x 106)3

    Magnetic materials used forconstruction of electromagnets havelarge flux density with smallmagnetising field & low hysterisis loss.eg. soft iron.

    Q24.Interstellar space has a very weak M.Fof the order of 10-12 T. What is thesignificance.

    Ans. Interstellar space has very weakmagnetic field. Such fields bendscharged particles such as cosmic rays.Over small distances this bending is notdetectable. But at large distances thedylection can affect the passage ofcharged particles.

    Q25. We can produce induced field insidea material by keeping it in external M.F.The magnetic properties of a materialis stadied by defining the following.

    a) What is meant by (i) Magnetizing field(ii) Intensity of magnetization.

    b) Obtain the relation betweensusceptibility and permeability.

    c) What is polestrength of a bar magnethaving a magnetization 240Am-1? Thelength of the magnet is 0.2m and areaof cross section is 2cm2.

    Ans. a) (i) The current through a solenoidproduces a field called Magnetising field(H) given by H = ni = 0 H unit A/m

    (ii) The magnetic moment per unit volumeof a material placed in a magnetisingfield is called magnetization.

    M =

    unit A/mb) Consider a solenoid carrying a current

    produces a magnetizing field H. A softiron piece placed in side it acquires amagnetization M.Then B = 0 (H + M)

    diving by H, = 0 (1 + )

    But = & =

    mv

    BH

    MHB

    HMH

    Aswathy Books 35

  • = 0 (1 + )

    = 1 +

    r = 1 +

    c) M = = =

    P = Ma M = 240 = 240 x 2 x 10-4 a = 2 x 10-4m2 = 480 x 10-4 Am

    Previous Questions1. The captain of a ship sailing in the

    Atlantic Ocean has to travel in the northdirection to reach the nearest port. Hefinds that the magnetic declination ofthe present position of the ship is 50 22East.

    a) What is meant by magnetic declinationb) The captain has a magnetic compass

    needle with him. By how much angleand in which direction should he deviatehis ship from the north direction, pointedby the magnetic needle to reach theport.

    c) Suppose magnetic needle is capableof rotating in a vertical plane about ahorizontal axes. If the ship reaches themagnetic pole of earth in whichdirection will be magnetic needle point.

    (March 2011)Ans. a) Magnetic declination at a place is

    the angle between magnetic meridianand geographic meridian at the place.

    b) 50 22 towards West.c) Vertical.Q2.A magnetic needle made of iron is

    suspended in a uniform external M.F.It experiences a torque and needlestarts oscillating.

    a) Write down the frequency of oscillationof the magnetic needle.

    b) If the magnetic needle is heatedbeyond curie temperature while it isoscillating, then its period a) increasesb) decreases c) remains the same d)becomes infinity. (March 2010)

    0

    mv

    P x 2la x 2l

    Pa

    Ans. T = 2

    f =

    b) Increases. Beyond the curietemperature the magnetic needlebecomes paramagnetic and its momentm decreases.

    3. How will you classify diamagnetic andparamagnetic materials based onmagnetic susceptibilities?

    (March 2009)Ans. For diamagnetic substances is small

    and negative.For paramagnetic substances is smalland position.

    Q4. Match the following.

    Feriomagnetism < 0 Aluminium, Calcium

    Paramagnetism >>0 Lead Bismuth

    Diamagnetism >0 Iron, Cobalt (March 2011)

    Ans.

    Feriomagnetism >> 0 Iron, CobaltParamagnetism >0 Aluminium,

    Calcium

    Diamagnetism

  • Ans: a) Diamagnetic b) Ferromagnetic c) Diamagnetic d) Paramagnetic

    Q6.An iron rod of 0.1m2 cross section issubjected to a mangnetising field of1000Am-1. Calculate the magneticpermeability. Given susceptibility of iron= 599.

    Ans: = 0(1 + m)= 4 x 10-7 (1 + 599)= 4 x 10-7 x 600= 7.54 x 10-4TmA-1

    Q7.Draw the hysterisis loop for soft iron andsteel.

    Ans:

    Q8. If a toroid uses Bismuth as its core,will the field in the core be slightlygreater or less than when the core isempty?

    Ans: Slightly less, since bismuth isdiamagnetic.

    Q9. Define Bohr magneton. What is itsunit.

    Ans. It is the magnetic dipolemomentassociated with an atom due to orbitalmotion of an electron in the first orbit ofhydrogen atom.1 Bohr magneton, B = 9.27 x 10-24 Am2

    Q10. A paramagnetic substance ofsusceptibility 3 x 10-4 is placed in amagnetising field of 4 x 10-4 am-1.Calculate (i) Intensity of magnetization

    soft iron

    steel

    H

    M

    (ii) Relative permeability.Ans. Given m = 3x10-4, H = 4 x 10-4 Am-1

    I = m H = 3x10-4 x 4 x 104 = 12 Am-1

    r = + 13 x 10-4 + 1 = 1.0003Q11. Name the line joining place having

    same (i) Declination (ii) Dip (iii)Horizontal intensity.

    Ans. (i) Same declination Isogonic line(ii) Same dip Isoclinic line(iii) Same horizontal intensity

    Isodynamic line

    Aswathy Books 37

  • 6. ELECTROMAGNETIC INDUCTION

    Q1.Electromagnetic induction was firstdiscovered by Faraday. The directionof this emf is given by Lenz.

    a) State the laws of eletromagneticinduction.

    b) How will you find the direction ofinduced emf, using this law.

    c) Flux in a closed circuit of resistance12 varies with time t as = 9t2+ 6t + 3Milli Webber where t is in seconds.Find the emf and current induced in thecoil in t = 3 s.

    Ans.a)(i) Faradays law states that themagnitude of the induced emf is equalto the rate of change of magnetic fluxlinked with it. =

    ii) Lenzs law, The direction of the inducedemf is such that it opposes the causeproducing the change. = -

    b) When a coil is brought near a north poleof a magnet, the magnetic flux linkedwith the coil begins to increase. Thecoil generates an induced emf so as tooppose the increase in flux. This ispossible if the coil can oppose the northpole of the magnet. To oppose the faceof the coil near to the magnet shows anorth polarity. Similarly if the north poleis moved away from the coil the face ofthe coil shows south polarity whichattracts the north pole. Thus one candetermine the direction of inducedcurrent in a coil.

    c) = -

    = - (9t2 + 6t +3)

    = -18t - 6

    ddt

    ddt

    ddtddt

    When t = 3 sec = -18 x 3 - 6 = - 54 - 6 = -60v

    I = = = 5 A

    2. The north pole of a magnet goes to andfro and then away from the coil.

    a) What is the direction of the inducedcurrent in the coil in each case.

    b) Define self induction & inductance.c) A coil has an inductance of 0.03 H.

    Calculate the emf induced when currentin the coil changes at a rate of 200 A/S

    Ans. a) When the north pole of the magnetgoes into the coil, the induced currentflow it such as to make the face of thecoil acquire north polarity. So currentin the coil is viewed in anticlock wisedirection.

    When the north pole is movedaway from the coil by Lenzs law theface of the coil acquire south polarityand the direction of the induced currentis clockwise.

    b) The property of the coil to oppose thegrowth and decay of current through itis called self induction I = L IWhen I = 1 A1 = LInductance is the flux linked with a coilwhen unit current is passed through it.

    c) = -

    = -L

    = 0.3 x 200 A/S

    = -60 v

    R

    6012

    S N

    ddt

    dIdt

    Aswathy Books 38

  • Q3. The plot of magnetic flux () versescurrent I is shown in fig. for twoinductors A and B. Which of the twohas larger value of self induction?

    Ans. As L = /I, therefore L for A is greaterthan L for B.

    Q4. A train is moving with uniform speedfrom north to south. Will any inducedemf appear across the ends of its axis?

    Ans. Yes, it will appear as the train isintercepting vertical components ofearths M.F.

    Q5. Why are oscillations of copper sheetin a magnetic field highly damped?

    Ans. This because of eddy currentsdeveloped in the copper sheet.

    Q6. Explain why coils are usually doublewound.

    Ans. The resistance coils are doublewound be avoid inductance effect. MFdue to current in one hay of the coil iscancelled by the MF due to current inthe other half of the coil.

    Q7. How is the mutual inductance betweena pair of coils affected when.

    i) Seperation between the coils.ii) The number of turns of each coil is

    increased ?iii) A thin iron sheet is placed between the

    two coils, other factors remaining thesame. Explain your answer in eachcase.

    Ans.(i) When the relative distance betweenthe coil is increased, the leakage of fluxincreases which reduces the magneticcoupling of the coil. So magnetic fluxlinked with all the turns decreases.

    I

    B

    A

    Hence mutual inductance will bedecreased.

    (ii) Mutual inductance for a pair of coil isgiven by M = K L1L2. when the No.of turns in each coil increases, mutualinductance also increases.

    (iii) When thin sheet is placed between thetwo coils, the mutual inductancedecreases due to the opposite eddycurrent set up in the iron sheet. As therate of change of magnetic flux in thecoil decreases, induced emfdecreases.Hence mutual inductancedecreases.

    Q8. How does the self inductance of an aircore coil change when (i) the numberof turns in the coil decreased (ii) an ironrod is introduced in the coil.

    Ans.(i) When the No. of turns in the coil isdecreased, self inductance decreases.

    (ii) If core of a magnetic material of relativepermeability r is introduced.L| = r .L, Self inductance increases.

    Q9.Two concentric circular loop, one ofsmall radius r1 and the other of largeradius r2 such that r1

  • place a light metallic disc at the top ofit. When the current is switched onthe disc is thrown up in the air.

    a) Why does the disc go up.b) How is the repulsive force produced.Ans. a) Induced currents are produced in

    the disc which act as a manner that itopposes the increase in magnetic fluxthrough it. As a result the disc is thrownup.

    b) When the circuit is switched on, thecurrent starts growing through thesolenoid. Hence magnetic flux throughthe coil increases from zero to somefinite value. Induced currents areproduced in the coil which oppose theincrease in magnetic flux through it.Hence the disc is thrown up.

    Q11. A small resistor is usually put inparallel to the current carrying coil ofan electromagnet. What purpose doesit serve?

    Ans. This prevents spark at the openswitch produced by the self inducedcurrent at the break. These sparks maydamage the insulation.

    Q12.A boy uses a coil of wire (90m long)without unwinding it to light a bulb using230 v mass. Then he anwinding it andagain lights the bulb.

    a) What difference does he observe?b) Name the phenomenon in the first

    case.Ans. Since the coil is wound in the first

    case, it has self inductance. Since acis used induced back emf is producedin the coil. Hence the bulb glows dim.But in the second case inductanceeffect is negligible. So the bulbbecomes brighter.

    Q13.a) When a magnet falls through avertical coil, will its acceleration bedifferent from the acceleration due togravity.

    b) Why does the acceleration of a magnetfalling through a long solenoiddecrease?

    Ans. a) Yes, This is because the motion ofthe magnet will be opposed inaccordance with Lenzs law.

    b) This is due to the opposition offered bythe induced emf.

    Q14. Name the physical quantity whose SIunit is weber. Is it a scalar or a vertorquantity?

    Ans. Magnetic flux. It is a scalar quantity.

    Q15. Find the induced current in the circularloop KLMN of radius r1 if the straight wirePQ carreis a steady current of I ampere

    Ans.Since the current is steady, themagnetic flux linked with the loop shallnot change. So there will be no inducedemf.

    Q16. Why birds fly off a high tension wirewhen current is switched on.

    Ans. When current begins to increase fromzero to maximum value, a current isinduced in the body of the bird. Thisproduces a repulsive force and the birdflies off.

    Previous Questions

    1. Sometimes when we switch off theelectrical equipments, a spark is seenwithin the switch. This is explained onthe basis of electromagnetic induction.

    a) What is inductionb) Current in a circuit falls from 5A to 0A in

    0.1s. If an average emf of 200v isinduced, determine the self inductanceof the circuit.

    (March - 2011)a) Question No. 2(b)

    >P Q

    N L

    M

    K

    I

    Aswathy Books 40

  • b) = 200v, dI = 0-5 = -5A, t = 0.1

    = -L

    L = - = -200 x = 4 H

    Q2.A jet plane is travelling west at a speedof 1800 km/hr. What is the voltagedifference between the ends of thewings 25m long if the earths magneticfield at the location has a magnitude of5 x 10-4 J and the dip angle is 300?

    Ans:Induced emf between the ends of thewings is = Bl vSince the plane is flying horizontally itswing moves perpendicular to thevertical intensity at the place.B = Vertical intensity = 7 Sin

    = 5 x 10-4 x Sin 30= 5 x 10-4 x .5= 2.5 x 10-4J

    I = 25m, v 1800km/hr = 1800 x m/s = 500m/s = Bl v = 0.25 x 10-4 x 25 x 500 = 3.125v

    Q3.Figure shows a bulb connected in anelectrical circuit.

    a) When the key is switched on the bulbattains the maximum glow after a shortinterval of time. Which property of thesolenoid is responsible for the decay.

    b) The flux linked with a solenoid changesfrom 0 to 1wb in 2s. Find the inducedemf in the solenoid.

    c) If 3v battery is replaced by ac sourceof 3v with the key closed, what will beobserved? Justify your answer.

    dIdtdtdI

    0.1

    -5

    518

    Ans:a) Self Inductionb) d = 1 - 0 = 1wb dt = 2s

    e = = = 0.5v

    c) Brightness decreases since the currentthrough the solenoid always changes,there is a back emf in the circuit. So theeffective emf decreases and so thebrightness of the bulb decreases.

    Q4. What is the self inductance of a solenoidof length 0.4m, area of corss section 20x 10-4 m2 and total number of turns 800?

    Ans. L =

    = 4.023 x 10-3 HQ5. Why a thick metal plate oscillating about

    a horizontal axis stops when a strongmagnetic field is applied on the plate?

    Ans.This because eddy currents areproduced and this opposes themechanical motion.

    Q6. A train is moving with uniform velocityfrom north to south. Will any inducedemf appear across the ends of the axle.

    Ans. Yes, the vertical component of earthsmagnetic field shall induce emf.

    Q7. Name the physical quantity whose SIunit is Weber. Is it a scalar or a vectorquantity?

    Ans.The Physical quantity is magnetic flux.It is a scalar quantity.

    Q8. When a fan is switched off, a spark isproduced in the switch. Why?

    Ans. At the time of break of the circuit, alarge emf is induced which opposes thedecay of current in the circuit. Thisionises the air between the contacts ofthe switch. Consequently spark isproduced.

    Q9. Why birds fly off a high tension wirewhen current is switched on?

    Ans.When current begins to increase fromzero to maximum value, a current isinduced in the body of the bird. Thisproduces a repulsive force and the birdflies off.

    d dt

    12

    ( )

    ON2Al4 x10-7 x 8002 x 20 x 10-4

    0.4

    Aswathy Books 41

  • 7. ALTERNATING CURRENT

    Q1. Write the general equation for theinstantaneous emf of a 50Hz generatorwhose peak voltage is 250Hz.

    Ans. Eo = 250v = 50 Hz w = 2 = 2 x 50 = 100

    E = Eo sin wtE = 250 sin 100 t

    Q2. The instantaneous current from an ACsource is I = 6 sin 314t. What is therms value of the current.

    Ans. Comparing with the equationI = I0 sin wtI0 = 6 A

    Irms = = = 4.24 A

    Q3. What is the SI unit ofAns. ohm

    Q4. Why a DC voltoneter and DC ammetercannot read AC?

    Ans. This is because the average valuevalue of AC over a complete cycle iszero.

    Q5.A transformer has an efficiency of 80%& work at 100v and 4 KW. If thesecondary voltage is 240v, claculate thePrimary & Secondary currents.

    Ans. = 80% Vp = 100v,Vp Ip = 4 x 103 W

    Ip = = 40 A

    =

    0.8 =

    Is =13.33 A

    I02

    6

    21

    c

    4 x 103 W100 v

    Vs IsVp Ip

    240 x Is4 x 103

    Q6. A town situated 20 Km away from apower plant generating power at 440v,requires 600 KW of electric power at220v. The resistance of the two wirecarrying power is 0.4 per km. Thetown gets power from the line through3000v - 220v step down transformer ata substation town.

    (i) Find the line power losses in the formof heat.

    (ii) How much power must the plant supplyassuming there is religible power lossdue to leakage?

    Ans. Total length of the wire = 2 x 20km =40km

    Resistance of the wire, R = 40x0.4 = 16Irms = = A = 200 A

    (i) Power loss in the form of heat= I2 rms R = 2002 x 16 = 64 x 104 W

    = 640 KW(ii) Power supplied by the plant =

    power demand + power loss =600 + 640 = 1240 KW

    Q7. An inductor of inductance 5mH isconnected in series with a capacitor ofcapacitance 20F and a variablefrequency sinusoidal supply as shown.

    a) When the frequency of the supply is250Hz, the rms current in the circuit is230 mA. calculate

    (i) The reactance of the inductor.(ii) The rms voltage across the inductor(iii) The rms voltage across the capacitor.b) By reference to the current in the inductor

    and the capacitor. State the phase ofp.d across

    PV

    600 x 103

    3000

    . .~5 mH 20PF

    Aswathy Books 42

  • (i) The inductor(ii) The capacitor.Ans. a) L = 5 mH = 5 x 10-3H

    C = 20F= 20 x 10-6Ff = 250 HzI = 230 mA .23A

    i) XL = Lw = L x 2 f = 5 x 10-3 x 2 x3.14 x 250 = 7.85

    ii) Vrms = Irms x L = .23 x 7.85 = 1.81 V

    iii) Vrms = Irms x

    = .23 x

    = 7.33 Vb) i) P.d leads the current by a phase /2ii) P.d lags by a phase /2Q8.Outline the function of a capacitor in

    an ac circuit.Ans. Uses of a capacitori) To produce a phase difference of

    /2 between current and voltage.ii) To block dc and to allow high frequency

    ac to pass through a circuit.

    Q9. A cellular phone transmits andreceives electromagnetic waves offrequencies between 806 and 902MHZ. Why do cellular phonessometimes have poor reception insidesteel frame office building.

    Ans.Electromagnetic waves arriving fromoutside the building cannot propagateinto conductors like steel. Insteadwaves are reflected back, outside. Soless wave energy reaches the phone.

    Q10.A filament lamp and a condenser withair between the plates are connectedin series to an ac source.What happensto the brightness of the lamp when adielectric slab is inserted between theplates of the capacitor? Explain.

    Ans. Capacitive reactance XC =

    decreases, the current in the circuitincreases and brightness increases.

    1C 1

    20 x 10-6 x 2 x 3.14 x 250

    12fc

    Q11. Explain why the transmission ofelectrical energy in national distributionsystem is carried out with alternatingcurrent and with a high voltage.

    Ans. AC can be stepped up or downconvieniently without much power lossby a transfer. High voltage is used toreduce transmission loss.

    Q12. What is the relation between qualityfactor and selectivity of the circuit.

    Ans. Quality factor is a measure ofsharpness of resonance or selectivity.The greater the quality factor, greateris the selectivity.

    Q13. The divisions marked on the scale ofan ac ammeter is not equally spaced.Why?

    Ans.An ac ammeter is constructed on thebasis of heating effect of electrticcurrent. Since heat produced varies asthe square of the current, the divisionsmarked on the scale are not equallyspaced.

    Q14. What is the difference betweenresistance and reactance withreference to an ac.

    Ans. An ohmic resistance has a fixed valuefor resistance while reactance changeswith frequency of ac supply.

    Q15. Which is more dangerous to use acor dc. Why?

    Ans. Ac is more dangerous than dc of thesame voltage. It is because the peakvalue of ac is more than the rms value.

    Q16. A capacitor blocks dc but allows ac topass through it. Explain why?

    Ans.The capacitive reactance XC =

    For dc f = 0 so XC will be infinite. But achas a definite value for an ac. Henceac can pass through a capacitor.

    Q17. Explain the importance of powerfactor.

    Ans. The average power P = IV E

    V Cos .In

    electric circuits operating some electric

    12fc

    Aswathy Books 43

  • appliances, the power factor is kept aslow as possible. It is done so that thepower consumption in the circuit is low.On the other hand when power istransmitted, the powerfactor is made aslarge as possible.

    Q18. A lamp is connected in series with acapacitor. Predict your observation fordc and ac connections. What happensin each case if the capacitance of thecapacitor is reduced?

    Ans. If dc source is connected to the bulbthrough a capacitor, the capacitor offers

    an infinite reactance (XC = = )

    and hence the lamp doesnot glow. Thedc source charges the capacitor, but thelamp will not glow as the capacitoroffers an infinite resistance to dc.

    If ac is applied the capacitor offersa finite reactance depending on thefrequency of ac. When capaciter isconnected the lamp glows. Thebrightness of the glow depends on thevalue of capacitance C of thecapacitor. Reducing C will increase XCand the lamp will shine less brightlythan before.

    Q19. The v - I values obtained from atransformes constructed by a studentis shown in this table.

    a) Identify the trasformer as step up orstep down.

    b) How much power is wasted by thetransformer.

    c) What are the possible energy lossesin a transformer.

    d) If the input voltage is 48V and inputcurrent is 1A is it possible to light 240V,100W electric bulb using the abovetransformer? Justify the answer.

    12fc

    Primary coil Secondary coilVoltage Current Power Voltage Current Power200V 1 A ......... 1000V .........1

    10 A

    Ans. a) Step up transformer.b) Input power - VI = 200 x 1 = 200 W

    Output power = VI = 1000x = 100WPower wasted = 200-100 = 100 W

    c) Iron loss, Copperloss, hysterisis lossand magnetic flux leakage.

    d) This is not an ideal transformer. Due tothe energy losses output power is lessthan the input power. There is 50% loss.Input power = 48 x 1 = 48W. So theoutput power will be less than 48W (only24W). But we require 100W output tolight 100W. So it is not possible to light240v, 100W electric bulb.

    Q20.A boy constructed a circuit to tuneradio signals as shown and received theprogramme from a near by radiostation.

    a) Which phenomenon is made use ofradio tuning.

    b) Calculate the frequency of the wave thatcan be clearly accepted by the circuitand impedence corresponding to thesame frequency.

    Ans.a) LCR resonance.b) L = 20 H = 20 x 10-6 H

    C = 0.1 F = 0.1 x 10-6 FR = 10

    f = =

    110

    >

    >

    ^

    vvv vvv

    llllll

    20H to receiver

    0.1 F

    10

    12Lc

    Aswathy Books 44

  • = 112.6 KHZZ = R = 10

    Q21.The given circuit diagram shown aseries LCR circuit connected to avariable frequency 230V source.

    i) Determine the source frequency whichdriver the circuit in resonance.

    ii) Obtain the impedance of the circuit andthe amplitude of current at theresonating frequency.

    iii) Determine the rms potential dropsacross the three elements of the circuit.

    iv) How do you explain the observationthat the algebraic sum of the voltagesacross the three elements obtained in(iii) is greater than the supplied voltage?

    Ans. i) Resonant angular frequency,

    w = =

    = 50 rad/sec.

    f = = Hz

    ii) The impedence of the circuit is givenby

    Z = R2 + ( L - ) 2

    At resonance L =

    Z = R = 40The current amplitude at resonance ie.

    i0 = = =

    1

    2 x 3.14 x 20 x 10-6 x 0.1 x10-6

    . .

    ~

    5 mH 40vvv

    80F

    lLc

    l

    5 x 80 x10-6

    w2

    25

    l

    c

    l c

    EoZ

    2 ErmsR

    1.414 x 23040

    = 8.13 AThe rms current is Irms =

    = = 5.75 A

    iii) The rms potential drop aross L, C & RareVL = irms x L

    = 5.75 x 50 x 5= 1437.5 V

    VC = irms x

    = 5.75 x

    = 1437.5 VVR = irms x R = 5.75 x 40

    = 230 Viv) The voltage across R, L and C are not

    in the same phase. Therefore theycannot be added like an ordinarynumber in LCR circuit, V = VR2 + (VL- VC)2

    Previous Questions

    Q1a) Fill in the blanks

    If W is the angular frequency of ac, thenthe reactance offered by inductance Land capacitance C are respectively, XL- ............... and XC = ..................

    b) An electric bulb B and a parallel platecapacitor C are connected in series asshwon in figure. The bulb glows withsome brightness. How will the glow ofbulb affected on introducing a dielectricslab between the plates of thecapacitor? Give reason in support ofyour answer.

    c) Give below are two electric circuits A &B. What is the ratio of the power factorof the circuit B to that of A?

    ErmsZ

    23040

    l c l

    5 x 80 x10-6

    ~

    CB

    ac

    Aswathy Books 45

  • (March 2012)

    Ans. a) XL = LW and XC =

    b) When dielectric slab introducedcapacitance C increases andcapacitive reactance XC =

    decreases. So current through thecircuit decreases and brightness of thebulb increases.

    c) For A, power facter Cos=

    =

    =

    = =

    For B, Cos =

    =

    =

    = =

    ~

    RXL = 3R

    ac

    vvv

    A

    ~

    RXC = 3R vvv

    B

    XC = R

    lcw

    lcw

    R

    R2 + X12RR2 + (3R)2

    R

    R2 + 9R2R

    10 Rl

    10R

    R2 + (XL- XC)2R

    R2 + (3R - R)2R

    R2 + 4 R2R5R

    l

    5

    =

    = =

    Q2.A series LCR AC circuit has greatpractical importance. It is used fortuning radio, T.V, Wireless set, etc.

    a) Obtain the expression for current in aseries LCR circuit using phasordiagram.

    b) Under what condition this circuit is usedfor tuning.

    (March 2011)Ans. a) In NCERT Text.b) To select a particular frequency using

    resistance principle.Q3. Seema constructed a series LCR circuit

    in the laboratory as shown in thediagram. She found that the voltageacross the inductor and capacitor areequal when the circuit is connected toan ac source.

    a) State the condition at which the voltageacross L and C becomes equal.

    b) Obtain the expression for the frequencyat which this situation occurs in a seriesLCR circuit.

    c) Find the voltmeter and Ammeterreadings in the circuit.

    (March 2010)Ans. a) Resonance

    b) L =

    2 =

    Cos A Cos B

    l

    10 l5

    l

    10x 5 l

    2

    ~

    R = 100vvv

    220 v, 50 Hz

    A 300v 300v v

    lC

    lLC

    Aswathy Books 46

  • =

    2 f =

    f =

    c) Ammeter reading I = =

    = 2.2A

    Voltometer reading V = IR = 2.2 x 100 = 220 V

    Q4.A radio can tune over the frequencyrange of a portion of MW broadcastband (800 KHz to 1200 KHz). If its LCcircuit has an effective inductance of200H. What must be the range of itsvariable condenses?

    Ans:L = 200H = 200 x 10-6Hf1 = 800KHz = 800 x 103 = 8 x 105Hzf2 = 1200 KHz= 1200x103 = 12 x 105 Hz

    f = f2 =

    c =

    For the lower frequency